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Question #75833

Difficulty: 8/10 | 🟢 Free

The Question

Let $l, m, n$ be the direction cosines of a line in three-dimensional space. Consider the equation $\Delta(x) = 0$ , where $\Delta(x) = \begin{vmatrix} x + l^2 & lm + \sqrt{241}n & ln - \sqrt{241}m \\ lm - \sqrt{241}n & x + m^2 & mn + \sqrt{241}l \\ ln + \sqrt{241}m & mn - \sqrt{241}l & x + n^2 \end{vmatrix}$ If $\alpha, \beta, \gamma$ are the roots of the equation $\Delta(x) = 0$ , find the value of $|\alpha^2 + \beta^2 + \gamma^2|$ .

Options

A.
481
B.
483
C.
241
D.
482

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Direction Cosines constraint|Scalar Triple Product Volume Logic|Factor Theorem in Determinants

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

The problem constructs a determinant representing the characteristic polynomial of a perturbed rank-1 matrix.

Students are trapped into a brutal algebraic expansion by the presence of the direction cosines and the obfuscating constant $\sqrt{241}$ .

The 'Golden Path' requires recognizing the matrix as $xI + \vec{u}\vec{u}^T + cK$ , where $K$ is the cross-product skew-symmetric matrix.

By applying matrix properties—specifically that $K\vec{u} = 0$ and $\text{adj}(K) = \vec{u}\vec{u}^T$ —students can elegantly find the eigenvalues without expanding.

The secondary trap tests if students correctly handle the squares of purely imaginary roots, where a sign error leads to the distractor 483.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Direction Cosines constraint
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free